respectively. On comparing with the well known formalism of SQM [1]},
we see that there is a supersymmetry in the problem and the scalar potential
\(\tilde{\phi }(x)\)  is just the superpotential of the Schrödinger formalism.
Further \(\tilde{\Psi }^{(1)}\)  and \(\tilde{\Psi }^{(2)}\) 
are the eigen functions of the
Hamiltonians \(H_{1}\equiv \hat{A^{\dagger }}\hat{A}\)  and
\(H_{2}\equiv \hat{A}\hat{A^{\dagger }}\)  respectively with \(\tilde{V}^{(1,2)}(x)\) 
being the partner potentials. Thus the eigenvalues and the eigen functions of
the two Hamiltonians are related except that one of them has an extra
bound state at zero energy so long as \(\tilde{\phi }(x \rightarrow \pm \infty )\) 
have opposite signs. Without loss of generality we shall always choose
\(\tilde{\phi }(x)\)  such that the ground state energy of \(H_1\)  is zero.
In that case the eigen functions and eigenvalues (\(\tilde{E}^{(1)}_{n}\)  and
\(\tilde{E}^{(2)}_{n}\) ) corresponding to these two Hamiltonians are related
to each other as follows [1]}
\(\tilde{\Psi }^{(2)}_{n}(x)=[\tilde{E}^{(1)}_{n+1}]^{-\frac{1}{2}}\hat{A}\tilde{\Psi }^{(1)}_{n+1}(x)\,,\) 
\(\tilde{\Psi }^{(1)}_{n+1}(x)=[\tilde{E}^{(2)}_{n}]^{-\frac{1}{2}}\hat{A}^{\dagger }\tilde{\Psi }^{(2)}_{n}(x)\) 
